Physics Solution Manual for 1100 and 2101

He starts a distance h above the top of the ramp

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Unformatted text preview: (1.7 m/s ) 2 Substituting this result into Equation (2) gives 2 m (1.7 m/s ) = 1 mv0 2 2 85. or v0 = 2 (1.7 m/s ) = 2.4 m/s 2 SSM WWW REASONING Gravity is the only force acting on the vaulters, since friction and air resistance are being ignored. Therefore, the net work done by the nonconservative forces is zero, and the principle of conservation of mechanical energy holds. SOLUTION Let E2f represent the total mechanical energy of the second vaulter at the ground, and E20 represent the total mechanical energy of the second vaulter at the bar. Then, the principle of mechanical energy is written for the second vaulter as 1 1 2 2 mv + mgh = mv20 + mgh 2 42f2442f 2 4424420 14 31 3 E2 f E20 Since the mass m of the vaulter appears in every term of the equation, m can be eliminated algebraically. The quantity h20 is h20 = h, where h is the height of the bar. Furthermore, when the vaulter is at ground level, h2f = 0 m. Solving for v20 we have 2 v 20 = v 2 f − 2 gh (1) 338 WORK AND ENERGY In order to use Equation (1), we must first determine the height h of the bar. The height h can be determined by applying the principle of conservation of mechanical energy to the first vaulter on the ground and at the bar. Using notation similar to that above, we have 1 1 2 2 mv1f + mgh1f = mv10 + mgh10 24 2 14 244 3 144244 3 E1f E10 Where E10 corresponds to the total mechanical energy of the first vaulter at the bar. The height of the bar is, therefore, v2 − v2 (8.90 m/s)2 − (1.00 m/s)2 h = h10 = 1f 10 = = 3.99 m 2g 2(9.80 m/s2 ) The speed at which the second vaulter clears the bar is, from Equation (1), 2 2 2 v20 = v2f − 2 gh = (9.00 m/s) − 2(9.80 m/s )(3.99 m) = 1.7 m/s ______________________________________________________________________________ 86. REASONING AND SOLUTION At the bottom of the circular path of the swing, the centripetal force is provided by the tension in the rope less the weight of the swing and rider. That is, 2 mv = T − mg r { FC Solving for the mass yields m= T 2 v +g r The energy of the swing is conserved if friction is ignored. The initial energy, E0, when the swing is released is completely potential energy and is E0 = mgh0, where h0 = r (1 – cos 60.0°) = 1r 2 is the vertical height of the swing. At the bottom of the path the swing's energy is entirely kinetic and is Ef = 1 mv 2 Setting Eo = Ef and solving for v gives 2 Chapter 6 Problems 339 v = 2gh0 = gr The expression for the mass now becomes 8.00 × 102 N T = = 40.8 kg 2 g 2 9.80 m/s 2 ______________________________________________________________________________ m= ( ) 87. REASONING The magnitude fk of the average kinetic friction force exerted on the skateboarder is related to the amount of work Wk the done by that force and the distance s he slides along the ramp according to Wk = ( f k cos θ ) s (Equation 6.1). Solving this equation for the kinetic friction force, we find fk = Wk (1) s cos θ The ramp is semicircular, so sliding from the top to the bottom means sliding one-fourth of the circumference of a circle with a radius of r = 2.70 m. Therefore, s = 1 ( 2π r ) = 1 π r . 4 2 Because the kinetic friction force is always directed opposite the skateboarder’s velocity, the angle θ in Equation (1) is 180°. Thus, Equation (1) becomes fk = To complete Wnc = the Wk 1 π r cos180 2 solution, o we =− 2Wk (2) πr will use the work-energy theorem ( 1 mvf2 − 1 mv02 ) + ( mghf − mgh0 ) (Equation 6.6) to find the work Wk done by the 2 2 nonconservative kinetic friction force. We can do this because the normal force exerted by the ramp, which is also a nonconservative force, is always perpendicular to the skateboarder’s velocity. Consequently, the normal force does no work, so the net work Wnc done by all nonconservative forces is just the work Wk done by the kinetic friction force: Wnc = Wk = ( 12 mvf2 − 1 mv02 ) + ( mghf − mgh0 ) = 1 m ( vf2 − v02 ) + mg ( hf − h0 ) 2 2 (3) SOLUTION Substituting Equation (3) into Equation (2), we obtain ( ) 2 2 1 m vf2 − v0 + mg ( hf − h0 ) 2 =− fk = − πr πr 2Wk (4) 340 WORK AND ENERGY The skateboarder falls from rest, so we have v0 = 0 m/s. He starts a distance h above the top of the ramp, which itself is a distance r above the bottom of the ramp. His vertical displacement, therefore, is hf − h0 = − ( h + r ) . The algebraic sign of this displacement is negative because he moves downward. Suppressing units, Equation (4) then yields ( ) 2 2 2 1 m vf2 − 0 − mg ( h + r ) 2 = − mvf − 2mg ( h + r ) = 2mg ( h + r ) − mvf fk = − πr πr πr (5) Thus, the average force of kinetic friction acting on the skateboarder as he slides down the ramp is fk = ( ) 2 ( 61.0 kg ) 9.80 m/s 2 (1.80 m + 2.70 m ) − ( 61.0 kg )( 6.40 m/s ) π ( 2.70 m ) 2 = 3.40 × 10 2 N CHAPTER 7 IMPULSE AND MOMENTUM ANSWERS TO FOCUS ON CONCEPTS QUESTIONS _____________________________________________________________________________________________ 1. (b) Kinetic energy, 1 mv 2 , is a scalar quantity and is...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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